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“How to destroy your rotator cuffs and stretch out your shoulder socket”

Well if you swim it with proper form, your shoulders should be fine. It’s just when you swim it sloppily that issues arise.

No way it’s just that repetitive motion that causes it. But I mean what do I know. I was a sprinter so I despised anything over 100yds

According to my coach (who is a very qualified and respected coach), proper form butterfly will not do any damage to your shoulders.

They have not announced what the future planes will be called. They have played with names (Dreamliner, Sonic Cruiser) but they could also simply move to higher numbers. One of the great things about numbers is that there's an infinite amount of them.

This is true, but “Boeing 299,792,458” doesn’t have quite the same ring to it.

They've already made far more than 10 types of airplanes. So, I guess they'd just keep making them.

But how do they name them then?

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Edit: it appears that the query got messed up upon submission, so just copy past this into the search bar:

(100w/69kg)/(power output of the sun/mass of the sun)

Thanks!

Comment deleted10 days ago

I’m actually just a Canadian watching from the sidelines, and was legitimately curious as to what the space force could offer for the future of space exploration because that’s an international subject. But hey thanks for insulting me instead of just having a proper conversation on the subject. It’s no wonder people think your party is a joke.

Edit: I apologize for assuming you were an American liberal. It was wrong of me to do. I’m sorry.

Edit edit: in his original post before he edited it, he said something along the lines of “but hey, let’s face it. If it was Obama announcing it, you’d have wet your bed in excitement.”

3rd and final edit: I realized that my remarks were wrong to make right after I made them. I didn’t delete them because people need to see why I’m apologizing.

Comment deleted9 days ago

Nice recovery there.

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Hey that kinda worked it slowed down for a while but eventually started but again!

Try imagining it connecting to two other gears in a triangle formation, such that it is touching two gears that are also touching each other. It would be impossible for any of them to spin without breaking (which I find doesn’t happen in these mind images).

They just move through each other.

Ahhh shit. The only other thing I could suggest would be imagine shooting it with a really heavy projectile repeatedly.

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HypoG1 commented on •

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So there’s no way I can really eli5 this and give you enough detail to do computation. But, I’ll try to make it as clear as possible.
Another word for the derived subgroup is the commutator subgroup, which I think makes more sense.
So, the derived subgroup of a group, denoted [G,G], is the *subgroup generated by the commutators of G*

What’s a commutator of G? Well, take any 2 elements g and h in G. Then their commutator is defined [g, h] = g^{{-1}} h^{{-1}} gh. One thing to note is this is only the identity if g and h commute, so be careful.

So to get the commutator subgroup of a group G, find the commutator of every pair of elements. (It’s not too hard if the group is small enough, and if you know the centre of the group it gets even easier. You also don’t need to check [g, g] since that’s obviously the identity. Then once you have every commutator, at worst, multiply them all together. Remember that the commutator subgroup is in fact a subgroup, so at worst it can be all of G, and it has to obey the usual restrictions on sizes for subgroups. It also turns out to be a normal subgroup.

Given that, you should be able to tell almost immediately what the commutator subgroup of S3 is without much work, since there are only 3 normal subgroups. I’m not going to give anything else away now, but if you need any other help, just let me know!

Also, a better place for this level of question might be r/learnmath if you ever need something of this level again.

Wow thank you! Very helpful. Do you count repeated elements when making the subgroup?

What do you mean by repeated elements? If you have an element [g, h] = [l,k] (so an element which is the commutator of more than one pair), you can only count it once.

If you mean, do you include the commutator of [g, g], You do include *every* commutator of G. Once you compute [g,g] for any element g, you should see why it’s important.

Sorry for the confusion - I meant the first thing you said. I was confused about it because I wrote out all the commuters of S3 on a piece of paper and got (1,2,3), (2,3,1), (3,1,2) repeated several times. Thanks !

It's a group that describes all of the ways you can arrange n different objects. So assume you have 7 apples sitting in front of you, and you take little pieces of paper with the numbers 1 through 7, and put them in front of the apples. This means that each apple is now moved to the new position on the piece of paper, so if you put number 5 in front of apple 1, it moves to position 5, whereas apple 5 moves to whatever number is on the piece of paper in front of it. If you have number 3 in front of apple 3, the apple doesn't move. This entire constellation of "move 1 to 5, 5 to 2, 3 to 3, ...." is the element of the Symmetrical group.

Specifically, in the way it's usually represented, you put in a number, and receive a different number based on the rearrangement. So say you are looking at S3, that's all rearrangements of 3 objects. This one then includes things like a rearrangement that switches 1 and 3, so it's a function s(x) with s(1)=3, s(2)=2, and s(3)=3. You can also create more complex cycles, like the one that pushes each number ahead by one space, so s(1)=2, s(2)=3, s(3)=1.

The value of "n" determines how many numbers you have to rearrange, and it always includes all functions that map each of those numbers onto one of those numbers, without using the same target number twice (because 2 objects can't move to the same position).

You can combine these rearrangements with each other, which is why they form a mathematucal group. So if you perform 2 rearrangements after each other, you can always describe the final position with just one rearrangement. Example, if you first switch 1 and 3, then move each apple 1 place ahead, like in the above example, you will end up with apple 1 on position 1, apple 2 on position 3, and apple 3 on position 2, so the result of those 2 rearrangements is the same as simply switching apple 2 and apple 3.

Thank you so much! Lifesavers.

Hey, I sent this to you as part of our message thread, but just in case anyone else wants to see it:

Okay, so we can start with symmetric groups. Let's say you have an ordered "tuple" of n things. We'll just use 3 for brevity. We can use symbols as the things:

(a, b, c)

A *permutation* over n things is just a function whose input is a tuple of size n (or an n-tuple) and whose output is another n-tuple where all the same symbols are there, but they can be in a different order.

The most basic permutation is the identity - given a tuple (a, b, c) it returns (a, b, c). Another permutation might reverse the order:

(a, b, c) -> (c, b, a).

There are n! possible permutations for an n-tuple. For 3, we have 3! = 6 permutations:

(a, b, c) - the identity

(a, c, b)

(b, a, c)

(b, c, a)

(c, a, b)

(c, b, a)

There are lots of ways to write permutations. Here's an example of a way. We can just write a list of numbers that shows what position each element ends up at. So the identity permutation on 3-tuples is [1, 2, 3] because the first element (a) goes to 1, and so forth. Using this notation, the list above is:

[1, 2, 3]

[1, 3, 2]

[2, 1, 3]

[3, 1, 2]

[2, 3, 1]

[3, 2, 1]

So for the symmetric group. Recall that a group is just a collection of elements, with an operation (+), that satisfies the rules of being a group. These rules are

closure: if X and Y are in the group, X+Y is in the group

associativity: (X+Y)+Z = X+(Y+Z)

identity: the group has an element E such that X+E = E+X = X for all X in the group

inverse: Each element X has an element Y (sometimes written X^{-1}) such that X+Y = Y+X = E

We can create a group *out of these 6 permutations* (for n = 3). The "+" operation is *composition*, which just means X + Y says you permute the list using Y first, then using X. So if we want to figure out what

[2, 1, 3] + [3, 2, 1]

is, we just do both permutations (starting with the one on the right). Starting with (a, b, c), we have

[3, 2, 1] maps (a, b, c) to *(c, b, a)*.

[2, 1, 3] maps *(c, b, a)* to (b, c, a)

(Italics just to show you that we took the output of the first permutation and used it as the input to the second one.)

So [2, 1, 3] + [3, 2, 1] is just the permutation that sends (a, b, c) to (b, c, a). Well we already know this one: we called it [3, 1, 2].

You can do the same thing for any pair of permutations f and g. f + g is just whatever permutation you end up with when you apply g to the tuple, then apply f to the resulting tuple.

I'll leave it up to you to convince yourself that this is really a group. You would need to show that the 4 group axioms (listed above) are satisfied. But it turns out they are.

So that's all the symmetric group is: it's a group of *permutation* elements which re-order tuples, combined with the *composition* operator which combines permutations.

The alternating group of size n is a subset of the symmetric group of size n. It just contains all the permutations on n items which have a special property.

The property is that if you do the permutation, then count how many pairs of symbols are "out of order", this count is even. Let's look at an example on 4-tuples:

The permutation resulting in (b, a, d, c) is even, because

*a and b are out of order*

a and c are in order

a and d are in order

b and c are in order

b and d are in order

*c and d are out of order*

So two pairs are out of order, and that's an even number. Thus the permutation [2, 1, 4, 3] is even, and it is in the alternating group on 4 elements.

For 3 elements, the alternating group consists of 3 permutations (it's always half of the total possible permutations). Here are the resulting tuples from those 3 permutations:

(a, b, c) [0 pairs out of order]

(b, c, a) [2 pairs out of order: ac and ab]

(c, a, b) [2 pairs out of order: ac and bc]

(For bigger tuples, the symmetric group is more "complicated", in that it isn't always just "rotations" like we see here.)

It turns out that this subset of permutations is in fact a group, which again you may need to convince yourself of.

I know that's a lot but hopefully you can get comfortable with each step and build up to being comfortable with the end result. There's kind of no "short version" of these things, unfortunately.

Thank you ! Very very very helpful.

I can try to build it up from some examples. I'll assume you know what a field is. We'll start in **Q**, the field of rational numbers, since it's the classical example of a field.

**Preliminaries**

A polynomial with coefficients in **Q** is just a polynomial of the form

a_0 + a_1*x + a_2*x^{2} + ... a_n*x^{n}

where all the a_i coefficients are rational numbers.

We know that we can rewrite any such polynomial in terms of its roots:

(x - r_1)(x - r_2)...(x - r_n)

where some of the roots might be repeated.

Splitting fields concern whether or not these roots r_i are rational numbers (since we are starting with **Q**), and if not what fields they live in.

**Examples**

Let's look at a simple polynomial: x^{2} - 1. We know that this can be factored as (x - 1)(x + 1) and the roots are 1 and -1. These roots are rational, and that's pretty satisfying. The polynomial had rational coefficients and the roots are rationals.

Now let's look at x^{2} - 2. We know this one factors into (x - sqrt(2))(x + sqrt(2)). This is less satisfying. The roots are not rational numbers. They live outside of our original field.

What field do they live in? Well we know they live in **R**, the set of all real numbers. And they live in a bigger field than that, **C**, the set of all complex numbers.

But do they exist in any smaller fields than **R**? It turns out they do.

**Field Extensions**

A field *E* is an *extension* of a field *F* if *E* contains *F* and the field operations of *E*, when restricted to elements of *F*, give the same results as the field operations of *F* do.

That's a lot of words, but as an example the real numbers are an extension of the rationals, because they contain all the rationals and when we multiply/add rationals the answer doesn't change just because we are working in **R**. In the same way, **C** is an extension of **R** and also an extension of **Q**.

**Back to our Examples**

It turns out that we can make fields that are "between" the rationals and the reals by picking an irrational number and "extending" the rationals with it. Here's an example:

All numbers that can be written as a + b*sqrt(2), where a and b are rational numbers.

It turns out that this is in fact a field (using the regular operations), which you may want to prove to yourself. It clearly contains the rationals (whenever b = 0), and the operations still "work" on the rationals. So it is an extension of the rationals. We call it **Q**(sqrt(2)).

This field is bigger than the rationals, but it's smaller than the reals. For example, it does not include the numbers sqrt(3) or *e*. But check this out - the roots of x^{2} - 2 are all (both) in this field! Recall that the roots are sqrt(2) and -sqrt(2), both of which are in our extended field.

It turns out that **Q**(sqrt(2)) is the *smallest* field that contains all the roots of x^{2} - 2. There's no field extension of **Q** that contains the roots, but doesn't contain all of **Q**(sqrt(2)).

That makes **Q**(sqrt(2)) the *splitting field* of x^{2} - 2, by definition.

**Another Example**

What about x^{2} + 1? We know this can be factored as (x - *i*)(x + *i*), so again the roots are not in **Q**. They are in **C**, but again they are in a smaller field extension that is between **Q** and **C**: the field of all numbers a + b**i*, where a and b are rationals. This is different than **C**, because it doesn't include lots of complex numbers like sqrt(2) and pi + sqrt(2)**i*.

But this field, **Q**(*i*), is the splitting field of x^{2} + 1. It's big enough to contain all the roots of the polynomial, but no field is smaller than it and still contains those roots.

**Final Remarks**

It turns out that every polynomial with coefficients in **Q** has a splitting field. You might need to combine a bunch of extensions to get there, but there is some field that contains all the roots of your polynomial and is an extension field of **Q**. For example, if we take

x^{4} - x^{2} - 2 = (x - *i*)(x + *i*)(x - sqrt(2))(x + sqrt(2)),

the splitting field ends up being **Q**(*i*)(sqrt(2)), which is the same as **Q**(sqrt(2))(*i*), which is just all the numbers that can be written as

a + b**i* + c*sqrt(2) + d**i**sqrt(2), where {a,b,c,d} are all rationals.

That field contains the roots of x^{4} - x^{2} - 2, and any other field that contains all the roots of x^{4} - x^{2} - 2 will contain all numbers that can be written that way.

Just to pique your curiosity, it's kind of interesting that **C** is "bigger" than **Q**(*i*)(sqrt(2)), in that it contains everything in that field. But **R** is neither bigger nor smaller than that field. There are numbers in **R** that aren't in that field, like pi. There are numbers that are in that field, but aren't in **R**, like *i*.

You can see the same thing happen if you think about **Q**(sqrt(2)) and **Q**(sqrt(3)) and **Q**(*i*). None of these fields contains any of the other ones. **R** contains (is an extension of) the first two and **C** contains all three. As for **R** and **Q**(*i*), neither one contains the other one.

So yeah, that can happen :)

Edit: typo

Wow that was amazing! Thank you! Really cleared it up for me. You’re clearly very knowledgeable about this subject.

[removed]

mathematics

It’s called the trans Atlantic accent! It was a mix of what was the American and British accent at the time. Upper class children were taught it in school.

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I think I'd support mandatory life sentences for violent rapists. I'm sick of hearing about a guy getting 18 months for rape and then raping someone weeks after getting out so he can spend another two years in jail, etc.

It’s fun to think about “oh these people did something terrible so they should face insane punishments”, which I don’t disagree with. But is that really the best way to go about it? What if they could be rehabilitated and released back to the public, like a normal human? Instead of costing thousands of taxpayer dollars to hold them in prison all their life, they actually get to go out and help humanity. Some of the Scandinavian countries are implementing a system like this and it works quite well.

My childhood best friend and all her siblings (one being 2 years old!!) Were raped by their uncle. Not only that he raped 15 other children and they all testified against him. He got 8 years. How is this fair. All proven rapists should be sentanced to life.

It’s fun to think about “oh these people did something terrible so they should face insane punishments”, which I don’t disagree with. But is that really the best way to go about it? What if they could be rehabilitated and released back to the public, like a normal human? Instead of costing thousands of taxpayer dollars to hold them in prison all their life, they actually get to go out and help humanity. Some of the Scandinavian countries are implementing a system like this and it works quite well.

HypoG1 commented on •

But not out of the pool (swimming).

In open-water Endurance swimming, women actually tend to set better times than men.

Well that’s not entirely true. All but 1 member of the men’s field at the 2016 open water 10km swim at the rio olympics beat the first place finisher for women’s field.

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